Understandable StatisticsEighth Edition By Brase and BrasePrepared by: Lynn SmithGloucester County CollegeEdited by: Jeff, Yann, Julie, and OliviaChapter 8: Estimation

Section 8.1Estimating When Is Known

Focus PointsExplain the meaning of confidence level, error of estimate, and critical value.Find the critical value corresponding to a given confidence level.Compute confidence intervals for when is known. Interpret the results.

Statistics QuoteThere are three kinds of lieslies, damned lies, and statistics. Benjamin Disraeli

Assumptions About the Random Variable xWe have a simple random sample of size n drawn from a population of x values.The value of sigma, the population standard deviation of x, is known.If the x distribution is normal, then our methods work for any sample size n.If x has an unknown distribution, then we require a sample size n 30. However, if the x distribution is distinctly skewed and definitely not mound-shaped, a sample size of 50 or even 100 or higher may be necessary.

Point Estimatean estimate of a population parameter given by a single number

Examples of Point Estimates

Examples of Point Estimates is used as a point estimate for .

Examples of Point Estimates is used as a point estimate for .

s is used as a point estimate for .

Margin of Errorthe magnitude of the difference between the point estimate and the true parameter value

The margin of error using as a point estimate for is

Confidence LevelA confidence level, c, is a measure of the degree of assurance we have in our results.The value of c may be any number between zero and one.Typical values for c include 0.90, 0.95, and 0.99.

Critical Value for a Confidence Level, cthe value zc such that the area under the standard normal curve falling between zc and zc is equal to c.

Critical Value for a Confidence Level, cP( zc < z < zc ) = c

Area Between z.99 and z.99 is .99

Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90.

P(-z0.90 < z < z0.90 ) = 0.90

.90

Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90.

P(0< z < z0.90 ) = 0.90/2 = 0.4500

.4500

Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90.

P( z < z0.90 ) = .5 + 0.4500 = .9500

.9500

Find z0.90 such that 90% of the area under the normal curve lies between z-0.90 and z0.90.

According to Table 5a in Appendix II, 0.9500 lies exactly halfway between two area values in the table (.9495 and .9505).

Averaging the z values associated with these areas gives z0.90 = 1.645.

Common Levels of Confidence and Their Corresponding Critical Values

Level of Confidence, c

Critical Value zc

0.90, or 90%

1.645

0.95, or 95%

1.96

0.98, or 98%

2.33

0.99, or 99%

2.58

More Confidence Values

C Confidence Interval for An interval computed from sample data in such a way that c is the probability of generating an interval containing the actual value of

Confidence Interval for the Mean of Large Samples (n 30)

Confidence Interval for the Mean of Large Samples (n 30)

Confidence Interval for the Mean of Large Samples (n 30)

Confidence Interval for the Mean of Large Samples (n 30)

Confidence Interval for the Mean of Large Samples (n 30)

Confidence Interval for the Mean of Large Samples (n 30)

Create a 95% confidence interval for the mean driving time between Philadelphia and Boston.

Assume that the mean driving time of 64 trips was 6.4 hours with a standard deviation of 0.9 hours.

= 6.4 hours

s = 0.9 hours

c = 95%,

so zc = 1.96

n = 64

= 6.4 hours s = 0.9 hours Approximate as s = 0.9 hours. 95% Confidence interval will be from

= 6.4 hours s = 0.9 hours c = 95%, so zc = 1.96 n = 64

95% Confidence Interval:6.4 .2205 < < 6.4 + .2205 6.1795 < < 6.6205 We are 95% sure that the true time is between 6.18 and 6.62 hours.

Calculator InstructionsCONFIDENCE INTERVALS FOR A POPULATION MEANThe TI-83 Plus and TI-84 Plus fully support confidence intervals. To access the confidence interval choices, press Stat and select TESTS. The confidence interval choices are found in items 7 through B.

Example ( is known with a Large Sample)Suppose a random sample of 250 credit card bills showed an average balance of $1200. Also assume that the population standard deviation is $350. Find a 95% confidence interval for the population mean credit card balance.

Example ( is known)Since is known and we have a large sample, we use the normal distribution. Select 7:ZInterval.In this example, we have summary statistics, so we will select the STATS option for input. We enter the value of , the value of x and the sample size n. Use 0.95 for the C-Level.

Example ( is known)Highlight Calculate and press Enter to get the results. Notice that the interval is given using standard mathematical notation for an interval. The interval for goes from $1156.6 to $1232.4.

Section 8.1, Problem 3Diagnostic Tests: Plasma Volume Total plasma volume is important in determining the required plasma the overall health and physical activity of an individual. (Reference: See Problem 2.) Suppose that a random sample of 45 male firefighters are tested and that they have a plasma volume sample mean of x-bar = 37.5 mL/kg (milliliters plasma per kilogram body weight). Assume that = 7.50 mL/kg for the distribution of blood plasma.

(a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error?

(b) What conditions are necessary for your calculations?

(c)Give a brief interpretation of your results in the context of this problem.

Solution

Section 8.1, Problem 7Salaries: College Administrators How much do college administrators (not teachers or service personnel) make each year? Suppose you read the local newspaper and find that the average annual salary of administrators in the local college is x-bar = $58,940. Assume that is known to be $18,490 for college administrator salaries (Reference: The Chronicle of Higher Education).

(a) Suppose that x-bar = $58,940 is based on a random sample of n = 36 administrators. Find a 90% confidence interval for the population mean annual salary of local college administrators. What is the margin of error?

Section 8.1, Problem 7(b) Suppose that x-bar = $58,940 is based on a random sample of n = 64 administrators. Find a 90% confidence interval for the population mean annual salary of local college administrators. What is the margin of error?

(c) Suppose that x-bar = $58,940 is based on a random sample of n = 121 administrators. Find a 90% confidence interval for the population mean annual salary of local college administrators. What is the margin of error?

(d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease?

(e) Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increased, does the length of a 90% confidence interval decrease?

Solution